The relationship between modern agricultural production and chemical fertilizers is very close. Only scientific and rational application of chemical fertilizers can achieve the goal of high yield and stability, high quality and high efficiency, low cost and environmental pollution reduction. In production, many farmers buy fertilizer into the home, and do not convert between the amount of chemical fertilizer and the physical quantity. Even a few technicians will not calculate it, which affects the application effect of fertilizer in production. The following describes several calculation methods, for reference only:
First, the calculation of the amount of fertilizer
1. If a farmer purchases 1 pack of 40kg urea, the package is marked with N:46%; buy 1 pack of 50kg potassium sulfate, the package is marked with K2O: 50%, ask the 2 packs of fertilizer pure How many kg are the quantities?
Calculation formula: fertilizer weight kg × bag marked quantity
(1) The pure N amount of 1 pack of 40kg urea is: 40×46%=18.4(kg)
(2) 1 pack of 50kg potassium sulfate pure K20 is: 50×50%=25(kg)
2. If a farmer purchases a pack of 40kg compound fertilizer, the content of the package N, P2O5, and K2O is 15:10:5. How many kg of N, P2O5, and K2O are used?
Still calculated according to the above formula:
(1) The amount of pure N is: 40 × 15% = 6 (kg)
(2) The amount of scalar P2O5 is: 40 × 10% = 4 (kg)
(3) The amount of scalar K2O is: 40 × 5% = 2 (kg)
Second, the calculation of the amount of chemical fertilizer application
1. Calculation of the physical quantity of single fertilizer
If the technical scheme is designed, 1 mu of rice should be applied pure N=10kg, P2O5=3kg, K2O=4kg. How many kg of urea containing 46% of N should be applied in 1 mu of paddy field? How much kg of calcium containing P2O5=16%? K2O=50% potassium sulfate? How many kg?
The calculation formula is: the amount of fertilized material = (designed fertilization amount of ÷ fertilizer effective amount) × 100
Calculated as follows:
Should be applied with the amount of N 46% urea = (10 ÷ 46) × 100 = 21.74kg
Should be applied with P2O516% calcium = (3÷16) × 100 = 18.75kg
Potassium sulfate containing 50% K2O = (4÷50) × 100 = 8kg
2. Calculation of the physical quantity of compound fertilizer combined with simple fertilizer
Taking pure N12kg, P2O5=4kg and K2O=6kg per acre as an example, the actual dosage of elemental fertilizer and compound fertilizer was calculated.
The amount of compound fertilizer applied should be calculated by designing the least amount of fertilization, and then adding the other two fertilizers.
If the content of nitrogen, phosphorus and potassium indicated on a compound fertilizer bag is 13:5:7, then the amount of compound fertilizer applied to the plot should be (4÷5)×100=80kg.
At the same time, it is also indicated that the application of 80 kg of compound fertilizer with NPK content of 13:5:7 is equivalent to pure N (80×13) ÷100=10.4kg, P2O5 (80×) applied to the soil. 5) ÷100=4kg, K2O (80×7) ÷100=5.6kg.
Because the proportion of nutrients in compound fertilizer is fixed, it is difficult to meet the demand for various nutrients in different soils of different crops at the same time. Therefore, it is necessary to add simple fertilizer to supplement.
The calculation formula is: the amount of supplementary fertilization = (designed fertilization amount - applied fertilizer amount) 有效 the effective content of fertilizer to be applied
From the above calculation results, it is seen that the amount of phosphate fertilizer has been met, and nitrogen and potassium fertilizers need to be supplemented.
According to the design of fertilization amount of pure nitrogen 12kg, K2O = 6kg, it is necessary to increase the application: urea (12-10.4) ÷ 46% = 3.478kg, potassium sulfate (6-5.6) ÷ 50% = 0.8kg.
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